Intuition The big pictureA planet orbiting the Sun is a closed dance governed by two unbreakable bookkeeping rules : the total mechanical energy E E E angular momentum L ⃗ \vec{L} L central, conservative force . These two conservation laws are so strong that, together, they force the orbit to be a conic section (ellipse, parabola, hyperbola). Almost all of orbital mechanics is just squeezing consequences out of E = const E = \text{const} E = const L ⃗ = const \vec{L} = \text{const} L = const
A central force points along the line joining the two bodies and depends only on the separation r r r F ⃗ ( r ⃗ ) = f ( r ) r ^ . \vec{F}(\vec{r}) = f(r)\,\hat{r}. F ( r ) = f ( r ) r ^ . F ⃗ = − G M m r 2 r ^ \vec{F} = -\dfrac{GMm}{r^2}\hat{r} F = − r 2 GM m r ^
WHY angular momentum is conserved.
Torque about the central body is τ ⃗ = r ⃗ × F ⃗ \vec{\tau} = \vec{r}\times\vec{F} τ = r × F F ⃗ ∥ r ⃗ \vec{F}\parallel \vec{r} F ∥ r
τ ⃗ = r ⃗ × f ( r ) r ^ = f ( r ) ( r ⃗ × r ^ ) = 0 ⃗ . \vec{\tau} = \vec{r}\times f(r)\hat{r} = f(r)\,(\vec{r}\times \hat{r}) = \vec{0}. τ = r × f ( r ) r ^ = f ( r ) ( r × r ^ ) = 0 .
Since τ ⃗ = d L ⃗ d t \vec{\tau} = \dfrac{d\vec{L}}{dt} τ = d t d L d L ⃗ d t = 0 ⃗ ⇒ L ⃗ = const \dfrac{d\vec{L}}{dt}=\vec{0} \Rightarrow \vec{L}=\text{const} d t d L = 0 ⇒ L = const
Intuition What does constantL ⃗ \vec{L} L mean physically?
L ⃗ \vec{L} L r ⃗ \vec{r} r v ⃗ \vec{v} v ⇒ \Rightarrow ⇒ the motion is confined to a single plane . The orbit is flat. That's a free gift from angular momentum conservation.
WHY energy is conserved.
Gravity is conservative : the work it does depends only on start and end positions, not the path. We can define a potential energy U ( r ) U(r) U ( r ) F ⃗ = − ∇ U \vec{F} = -\nabla U F = − ∇ U
d d t ( 1 2 m v 2 + U ( r ) ) = 0 ⇒ E = 1 2 m v 2 + U ( r ) = const . \frac{d}{dt}\left(\tfrac12 m v^2 + U(r)\right) = 0 \Rightarrow E = \tfrac12 m v^2 + U(r) = \text{const}. d t d ( 2 1 m v 2 + U ( r ) ) = 0 ⇒ E = 2 1 m v 2 + U ( r ) = const .
The minus sign says you are in a potential well — you must add energy to escape.
E = 1 2 m v 2 − G M m r \boxed{E = \frac12 m v^2 - \frac{GMm}{r}} E = 2 1 m v 2 − r GM m
In the orbital plane use polar coordinates ( r , θ ) (r,\theta) ( r , θ ) radial and tangential part:
v 2 = r ˙ 2 + ( r θ ˙ ) 2 . v^2 = \dot r^2 + (r\dot\theta)^2. v 2 = r ˙ 2 + ( r θ ˙ ) 2 .
The angular momentum magnitude is L = m r 2 θ ˙ L = m r^2\dot\theta L = m r 2 θ ˙ r θ ˙ r\dot\theta r θ ˙ r r r θ ˙ = L m r 2 \dot\theta = \dfrac{L}{mr^2} θ ˙ = m r 2 L
r 2 θ ˙ 2 = L 2 m 2 r 2 . r^2\dot\theta^2 = \frac{L^2}{m^2 r^2}. r 2 θ ˙ 2 = m 2 r 2 L 2 .
Substitute into E E E
E = 1 2 m r ˙ 2 + L 2 2 m r 2 − G M m r ⏟ U eff ( r ) . E = \frac12 m\dot r^2 + \underbrace{\frac{L^2}{2mr^2} - \frac{GMm}{r}}_{\displaystyle U_{\text{eff}}(r)}. E = 2 1 m r ˙ 2 + U eff ( r ) 2 m r 2 L 2 − r GM m .
Definition Effective potentialU eff ( r ) = L 2 2 m r 2 − G M m r . U_{\text{eff}}(r) = \frac{L^2}{2mr^2} - \frac{GMm}{r}. U eff ( r ) = 2 m r 2 L 2 − r GM m . centrifugal barrier (repulsive, ∝ 1 / r 2 \propto 1/r^2 ∝ 1/ r 2 ∝ − 1 / r \propto -1/r ∝ − 1/ r
The area swept per unit time is d A d t = 1 2 r 2 θ ˙ = L 2 m = const \dfrac{dA}{dt} = \dfrac12 r^2\dot\theta = \dfrac{L}{2m} = \text{const} d t d A = 2 1 r 2 θ ˙ = 2 m L = const Equal areas in equal times — that is angular momentum conservation in disguise.
For a bound elliptical orbit of semi-major axis a a a only on a a a :
E = − G M m 2 a \boxed{E = -\frac{GMm}{2a}} E = − 2 a GM m
This gives the vis-viva equation (combine the two boxes, eliminate E E E
v 2 = G M ( 2 r − 1 a ) \boxed{v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)} v 2 = GM ( r 2 − a 1 )
Orbit type
E E E a a a shape
Circle/Ellipse
< 0 <0 < 0 finite > 0 >0 > 0
bound
Parabola
= 0 =0 = 0 ∞ \infty ∞ escape (just barely)
Hyperbola
> 0 >0 > 0 < 0 <0 < 0 flyby
Escape speed: set E = 0 E=0 E = 0 r r r 1 2 m v e s c 2 = G M m r ⇒ v e s c = 2 G M r \frac12 mv_{esc}^2 = \frac{GMm}{r} \Rightarrow v_{esc}=\sqrt{\dfrac{2GM}{r}} 2 1 m v esc 2 = r GM m ⇒ v esc = r 2 GM
Worked example Example 1 — Speed at perihelion fromE E E L L L E = − 2.0 × 10 9 J E=-2.0\times10^{9}\,\text{J} E = − 2.0 × 1 0 9 J L = 4.0 × 10 13 kg⋅m 2 / s L=4.0\times10^{13}\,\text{kg·m}^2/\text{s} L = 4.0 × 1 0 13 kg⋅m 2 / s m = 1000 m=1000 m = 1000 G M = 3.99 × 10 14 GM=3.99\times10^{14} GM = 3.99 × 1 0 14 Step 1: Find a a a E = − G M m 2 a E=-\frac{GMm}{2a} E = − 2 a GM m Why? a a a a = − G M m 2 E = − ( 3.99 × 10 14 ) ( 1000 ) 2 ( − 2.0 × 10 9 ) = 9.98 × 10 7 m a = -\frac{GMm}{2E} = -\frac{(3.99\times10^{14})(1000)}{2(-2.0\times10^{9})} = 9.98\times10^{7}\,\text{m} a = − 2 E GM m = − 2 ( − 2.0 × 1 0 9 ) ( 3.99 × 1 0 14 ) ( 1000 ) = 9.98 × 1 0 7 m Step 2: At perihelion r ˙ = 0 \dot r=0 r ˙ = 0 v = L m r p v=\frac{L}{mr_p} v = m r p L r p r_p r p v p 2 = ( L m r p ) 2 v_p^2=\big(\frac{L}{mr_p}\big)^2 v p 2 = ( m r p L ) 2 E = 1 2 m v p 2 − G M m r p E=\frac12 mv_p^2-\frac{GMm}{r_p} E = 2 1 m v p 2 − r p GM m Why? Turning points are where the whole speed is tangential — easiest place to use L L L 1 / r p 1/r_p 1/ r p r p r_p r p v p = L / ( m r p ) v_p = L/(mr_p) v p = L / ( m r p )
Worked example Example 2 — Vis-viva in actionEarth-orbit, a = 7000 a=7000 a = 7000 v v v r = 7000 r=7000 r = 7000 v 2 = G M ( 2 r − 1 a ) = G M ⋅ 1 r v^2 = GM\left(\frac{2}{r}-\frac1a\right) = GM\cdot\frac{1}{r} v 2 = GM ( r 2 − a 1 ) = GM ⋅ r 1 r = a r=a r = a Why this step? For a circle r = a r=a r = a 2 r − 1 a = 1 r \frac2r-\frac1a=\frac1r r 2 − a 1 = r 1 v = G M / r = 3.99 × 10 14 / 7 × 10 6 ≈ 7.55 v=\sqrt{GM/r}=\sqrt{3.99\times10^{14}/7\times10^6}\approx 7.55 v = GM / r = 3.99 × 1 0 14 /7 × 1 0 6 ≈ 7.55
Worked example Example 3 — Will it escape?A probe at r = 10 7 r=10^7 r = 1 0 7 v = 9 v=9 v = 9 v e s c = 2 G M / r = 2 ( 3.99 × 10 14 ) / 10 7 = 8.93 v_{esc}=\sqrt{2GM/r}=\sqrt{2(3.99\times10^{14})/10^7}=8.93 v esc = 2 GM / r = 2 ( 3.99 × 1 0 14 ) /1 0 7 = 8.93 9 > 8.93 9>8.93 9 > 8.93 E > 0 E>0 E > 0 it escapes (hyperbolic). Why? E = 1 2 m v 2 − G M m r E=\frac12 mv^2 - \frac{GMm}{r} E = 2 1 m v 2 − r GM m v > v e s c v>v_{esc} v > v esc
Common mistake "Kinetic energy is conserved in an orbit."Why it feels right: energy is conserved, and KE seems like 'the energy'. The fix: only the sum K E + U KE+U K E + U U U U
U = − G M m / r U = -GMm/r U = − GM m / r r → 0 r\to 0 r → 0 Why it feels right: U U U The fix: U U U E E E redistributed , not destroyed.
Common mistake "Angular momentum changes because speed changes around the orbit."Why it feels right: the planet clearly speeds up and slows down. The fix: L = m r 2 θ ˙ L=mr^2\dot\theta L = m r 2 θ ˙ r r r θ ˙ \dot\theta θ ˙ L L L
Common mistake "A more eccentric orbit has higher energy."Why it feels right: eccentric orbits look 'wilder'. The fix: E = − G M m 2 a E=-\frac{GMm}{2a} E = − 2 a GM m a a a not e e e a a a e e e
Recall Feynman: explain to a 12-year-oldImagine swinging a ball on a stretchy rubber band around your head. Two things never change no matter how the ball loops around: how fast it whips around times how far out it is (that's angular momentum — pull it in and it spins faster), and the total of its moving-energy plus its stretch-energy (that's total energy). Planets do exactly this with gravity instead of a rubber band. When a planet swoops close to the Sun it zooms fast; when it drifts far it crawls — but those two "savings accounts" always add up to the same total.
"LEAP" — L is fixed (orbit is flat, equal areas), E nergy fixed by A (semi-major axis), P erihelion is fastest. And "flat, fast, fixed" : motion is flat (constant L ⃗ \vec L L fast at perihelion , energy fixed by a a a
Recall Active recall checkpointCover the note. Can you (1) prove L ⃗ \vec L L U = − G M m / r U=-GMm/r U = − GM m / r E = − G M m 2 a E=-\frac{GMm}{2a} E = − 2 a GM m
Why is angular momentum conserved in a gravitational field? Gravity is central (
F ⃗ ∥ r ⃗ \vec F\parallel\vec r F ∥ r ), so torque
τ ⃗ = r ⃗ × F ⃗ = 0 \vec\tau=\vec r\times\vec F=0 τ = r × F = 0 , hence
d L ⃗ / d t = 0 d\vec L/dt=0 d L / d t = 0 .
What does constant L ⃗ \vec L L The motion stays in a single fixed plane (orbit is flat).
Derive the gravitational potential energy. U ( r ) = − ∫ ∞ r ( − G M m / r ′ 2 ) d r ′ = − G M m / r U(r)=-\int_\infty^r(-GMm/r'^2)dr' = -GMm/r U ( r ) = − ∫ ∞ r ( − GM m / r ′2 ) d r ′ = − GM m / r .
Write the total mechanical energy of an orbiting body. E = 1 2 m v 2 − G M m r E=\frac12 mv^2 - \frac{GMm}{r} E = 2 1 m v 2 − r GM m .
What is the effective potential and its two parts? U e f f = L 2 2 m r 2 − G M m r U_{eff}=\frac{L^2}{2mr^2}-\frac{GMm}{r} U e f f = 2 m r 2 L 2 − r GM m : centrifugal barrier (repulsive) + gravity (attractive).
State the vis-viva equation. v 2 = G M ( 2 r − 1 a ) v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right) v 2 = GM ( r 2 − a 1 ) .
What does the total orbital energy depend on? Only the semi-major axis:
E = − G M m 2 a E=-\frac{GMm}{2a} E = − 2 a GM m (independent of eccentricity).
Sign of E E E E < 0 E<0 E < 0 ellipse (bound),
E = 0 E=0 E = 0 parabola (escape),
E > 0 E>0 E > 0 hyperbola (unbound).
Derive escape speed from energy conservation. Set
E = 0 E=0 E = 0 :
1 2 m v 2 = G M m r ⇒ v e s c = 2 G M / r \frac12 mv^2=\frac{GMm}{r}\Rightarrow v_{esc}=\sqrt{2GM/r} 2 1 m v 2 = r GM m ⇒ v esc = 2 GM / r .
How does Kepler's 2nd law relate to L L L d A / d t = L 2 m = dA/dt=\frac{L}{2m}= d A / d t = 2 m L = const, so equal areas in equal times = constant angular momentum.
Where in the orbit is speed maximum and why? At perihelion;
r r r is smallest so
U U U is most negative, hence KE (speed) is greatest at fixed
E E E .
At a turning point (r ˙ = 0 \dot r=0 r ˙ = 0 v = L / ( m r ) v=L/(mr) v = L / ( m r ) since velocity is purely tangential there.
Kepler's Laws of Planetary Motion — 2nd law = angular momentum conservation; 3rd law links a a a Effective Potential and Orbit Classification — uses U e f f U_{eff} U e f f Vis-Viva Equation — direct child of energy conservation.Central Force Problem — general framework these laws come from.Escape Velocity and Hohmann Transfers — applications of E = 0 E=0 E = 0 a a a Conservative Forces and Potential Energy — why U U U
integrate force from infinity
E = half m v squared - GMm/r
L = m r squared theta-dot
Potential energy U = -GMm/r
Split v into radial and tangential
Intuition Hinglish mein samjhoDekho, jab koi planet ya satellite gravity ke under ghoom raha hota hai, toh do cheezein kabhi nahi badaltin — angular momentum L ⃗ \vec L L total energy E E E central force hai (hamesha center ki taraf), isliye uska torque zero hota hai — aur torque zero matlab L ⃗ \vec L L flat plane mein rehta hai, aur jab planet Sun ke paas aata hai toh tezi se ghoomta hai, door jaata hai toh dheere — yahi Kepler ka second law hai (equal areas in equal time).
Energy ki baat karein toh E = 1 2 m v 2 − G M m r E = \frac12 mv^2 - \frac{GMm}{r} E = 2 1 m v 2 − r GM m U = − G M m / r U=-GMm/r U = − GM m / r E E E r r r θ ˙ \dot\theta θ ˙ L L L
Sabse powerful result: E = − G M m 2 a E = -\frac{GMm}{2a} E = − 2 a GM m semi-major axis a a a par depend karti hai, eccentricity e e e vis-viva : v 2 = G M ( 2 r − 1 a ) v^2 = GM(\frac{2}{r}-\frac{1}{a}) v 2 = GM ( r 2 − a 1 ) E < 0 E<0 E < 0 E = 0 E=0 E = 0 E > 0 E>0 E > 0
